Count the Orthocenters

It's a common knowledge that three altitudes in a triangle intersect in a point known as the orthocenter of the triangle. So let's start with a triangle ABC and draw all three of its altitudes AHa, BHb, and CHc.

Now there is an interesting question: How many triangles with altitudes drawn are there in the diagram?

The answer is 4 and, besides ABC, there are three more triangles: AHB, BHC, AHC, H being the orthocenter of ABC. Indeed, since AHa is orthogonal to BC so BC is orthogonal to AH. (In general terms, the relation of orthogonality is symmetric and instead of claiming that one line is orthogonal to another we may simply say that the two lines are orthogonal.) Again, by the symmetry of orthogonality, AC serves as another altitude in ΔAHB. Thus, in ΔAHB we have three altitudes: CHc, BC, and AC that intersect at the point C, the orthocenter of ΔAHB.